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- Thread starter Adaggiote
- Start date

- May 25, 2007

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Just to be sure, must you memorize all the formulas for the MCAT, including the kinematics? Is there a pen and paper test available? Is all you have your. Are calculators allowed? If someone could educate me, that would be fantastic.

Thanks!

Pen, paper, and pencil is a thing of the past. The MCAT is on computer (CBT = computer based test). While you'd think it would be easier to have a calculator on the computer, they do not provide one nor are you allowed to use one.

Excess memorization is the kiss of death for many MCAT takers. They focus so much on memorizing facts and equations that they never fully learn the concepts and thought processes needed to apply them. If you do so many practice problems that equations stick, then you've done well. If you break out flash cards and memorize every kinematics equation, then come test time you'll find that you're not sure what they're asking and which equation to use.

Pen, paper, and pencil is a thing of the past. The MCAT is on computer (CBT = computer based test). While you'd think it would be easier to have a calculator on the computer, they do not provide one nor are you allowed to use one.

Excess memorization is the kiss of death for many MCAT takers. They focus so much on memorizing facts and equations that they never fully learn the concepts and thought processes needed to apply them. If you do so many practice problems that equations stick, then you've done well. If you break out flash cards and memorize every kinematics equation, then come test time you'll find that you're not sure what they're asking and which equation to use.Build your knowledge in context!

to clarify BRT's point, while the understanding the concepts is necessary to do well on the exam, you

Well, you don't actually have to memorize kinematic equations; you can always derive them. You always have to memorize some basic things like F = ma, but that should be something so ingrained in a scientist's mind that it shouldn't seem like a "formula" to memorize. Same goes for things like V = IR. To be sure, memorizing additional equations such as kinematic equations help, but don't let the memorization crowd out the understanding in your limited memory.

deriving something like d=vt +1/2at^2 during the mcat has got to be one of the least efficient ideas ever. it also involves calculus.Well, you don't actually have to memorize kinematic equations; you can always derive them. You always have to memorize some basic things like F = ma, but that should be something so ingrained in a scientist's mind that it shouldn't seem like a "formula" to memorize. Same goes for things like V = IR. To be sure, memorizing additional equations such as kinematic equations help, but don't let the memorization crowd out the understanding in your limited memory.

I think d = vt + 1/2at^2 is pretty basic math. But the reason I emphasize understanding over memorization is that a person who truly understands the material will naturally memorize or be able to derive any formulas needed. On the other hand, I run into plenty of people who memorize things without really understanding what they mean. Most times, they are unaware of when a formula applies (what assumptions have to be true).

there are some things you can/should derive. kinematics equations do not fall in this category.I think d = vt + 1/2at^2 is pretty basic math. But the reason I emphasize understanding over memorization is that a person who truly understands the material will naturally memorize or be able to derive any formulas needed. On the other hand, I run into plenty of people who memorize things without really understanding what they mean. Most times, they are unaware of when a formula applies (what assumptions have to be true).

- May 25, 2007

- 3,992

- 806

We're all saying the same thing for the most part.

If you do enough practice passages and questions to know how to apply the information, you have simultaneously memorized the information. You need knowledge for this exam, but it is best absorbed through practice as opposed to rote memorization.

And I have to side with bleargh on the idea that deriving equations on the MCAT is not the best use of time. Going from d = 0.5atexp2 + vot + do to vt = at + vo should be simple enough to do, but you should know the first equation from recall after having done enough problems and not try to derive it.

- Nov 6, 2007

- 8,810

- 163

- 37

- Resident [Any Field]

As others have said, though the test is computerized, you will be given a pencil and scrap paper. You will not be allowed to have a calculator.

It can be helpful to memorize some equations. I scrawled the basic kinematics and a few others in the corner of my scratch paper during the beginning of the test during the mini tutorial thing or whatever it is, just in case I had a brainfart during the test and made a stupid mistake later on. I also drew that little sine and cosine chart for 0, 30, 45, 60 and 90 degree angles.

And obviously, yes you need to understand the material and rote memorization isn't going to help you, but I'm prettttttty damn sure that having them in front of me already written before I got into the thick of the MCAT was a help.

It can be helpful to memorize some equations. I scrawled the basic kinematics and a few others in the corner of my scratch paper during the beginning of the test during the mini tutorial thing or whatever it is, just in case I had a brainfart during the test and made a stupid mistake later on. I also drew that little sine and cosine chart for 0, 30, 45, 60 and 90 degree angles.

And obviously, yes you need to understand the material and rote memorization isn't going to help you, but I'm prettttttty damn sure that having them in front of me already written before I got into the thick of the MCAT was a help.

- May 25, 2007

- 3,992

- 806

As ...

Not to hijack the thread and change course here, but I scanned your med apps Geekchick and all I can think is "are the adcom people at Jefferson insane or stupid?" How did they waitlist you with your well-rounded, star-studded profile? Amazing in every category! Maybe they did that to see how serious you are about their school, figuring you'd follow up and contact them if you were serious. I just don't get what they could be thinking.

- We now return you to your regularly scheduled thread

There ARE some that are worth knowing...

like:

Big Five Equation (Kinematics)

E = hf

E = mc^2

KE = 1/2*m*v^2

PE = mgh

etc...

But trust me, formulas have very little impact on how you actually perform. There will be questions that involve math, but will use equations that no prep book has ever taught, except TBR - TBR taught me to use unit-analysis for answers. This is really important.

like:

Big Five Equation (Kinematics)

E = hf

E = mc^2

KE = 1/2*m*v^2

PE = mgh

etc...

But trust me, formulas have very little impact on how you actually perform. There will be questions that involve math, but will use equations that no prep book has ever taught, except TBR - TBR taught me to use unit-analysis for answers. This is really important.

- Oct 16, 2007

- 4,197

- 36

- Medical Student

Ugh okay guys, kinematics lesson.

Deriving the kinematics equations from the definition of uniform acceleration__doesn't require calculus at all.__ **I recommend every MCAT student to derive them themselves so they can memorize them forever.** ** Of course do this before the MCAT.** It takes maybe 20 minutes.

vf = final velocity, vo = initial velocity

Let's start with the definition of acceleration: vf = vo + at (this one you have to learn by rote memorization). Since this is a linear equation, we can graph this. You should know that the slope of this graph is the acceleration and the area under the curve is displacement (d).

(1) vf = vo + at

Now let's figure out how we can use geometry to figure out the area to get displacement.

The area of the bottom rectangle is height*base or vo*t.

The area of each triangle is 0.5*height*base or 0.5*at*t or 0.5at².

The area of the overall rectangle (from the top red line to the time axis) is height*base or vf*t.

The area under the curve is therefore:

the bottom rectangle + the bottom triangle or

(2) d = vo*t + 0.5at²

the overall rectangle - the top triangle or

(3) d = vf*t - 0.5at²

Now what if we chop the top half of the triangle and put on the other side? We would get a rectangle whose area would be equal to the original area (see picture below).

The area of this rectangle is height*base or

(4) d = 0.5*(vo + vf)*t

The last equation is tricky. We have to plug in the first equation into the fourth equation. From equation (1), we rearrange to**t = (vf - vo)/a** and plug t into equation (4).

d = 0.5*(vo + vf)***(vf - vo)/a** or

(5) d = 0.5*(vf² - vo²)/a

Let's look at the MEANING behind each equation.

(1) vf = vo + at

This is the definition of acceleration. It means, that acceleration is equal to the change in velocity over time.

(2) d = vo*t + 0.5at²

What if we take out "0.5at²"? What does "vo*t" physically mean? That is just distance = rate * time for an object that is traveling at constant initial velocity. But since the object is accelerating, it will go an EXTRA distance. So we have to add the "0.5at²" in.

(3) d = vf*t - 0.5at²

"vf*t" would be the displacement if the object was traveling at the final velocity the entire time. But it didn't, it traveled at a slower velocity and accelerated. So we have to subtract some displacement because it wasn't going that fast the entire time. Thus, we have to subtract "0.5at²" out.

(4) d = 0.5*(vo + vf)*t

What is "0.5(vo + vf)"? That's the average velocity the object is traveling. So this equation really says "displacement = average velocity * time" which is an equation you already know.

(5) d = 0.5*(vf² - vo²)/a

This is just the restatement of equation (4). Multiplying the average velocity by time gives us the displacement. In this case, time = (vf - vo)/a which simplifies into this.

By understanding where the equations come from and what the meaning is behind each equation, you not only have an easier time remembering each equation, but you have the deeper understanding required to do well on the MCAT.

(Created with MS Paint)

Deriving the kinematics equations from the definition of uniform acceleration

vf = final velocity, vo = initial velocity

Let's start with the definition of acceleration: vf = vo + at (this one you have to learn by rote memorization). Since this is a linear equation, we can graph this. You should know that the slope of this graph is the acceleration and the area under the curve is displacement (d).

(1) vf = vo + at

Now let's figure out how we can use geometry to figure out the area to get displacement.

The area of the bottom rectangle is height*base or vo*t.

The area of each triangle is 0.5*height*base or 0.5*at*t or 0.5at².

The area of the overall rectangle (from the top red line to the time axis) is height*base or vf*t.

The area under the curve is therefore:

the bottom rectangle + the bottom triangle or

(2) d = vo*t + 0.5at²

the overall rectangle - the top triangle or

(3) d = vf*t - 0.5at²

Now what if we chop the top half of the triangle and put on the other side? We would get a rectangle whose area would be equal to the original area (see picture below).

The area of this rectangle is height*base or

(4) d = 0.5*(vo + vf)*t

The last equation is tricky. We have to plug in the first equation into the fourth equation. From equation (1), we rearrange to

d = 0.5*(vo + vf)*

(5) d = 0.5*(vf² - vo²)/a

Let's look at the MEANING behind each equation.

(1) vf = vo + at

This is the definition of acceleration. It means, that acceleration is equal to the change in velocity over time.

(2) d = vo*t + 0.5at²

What if we take out "0.5at²"? What does "vo*t" physically mean? That is just distance = rate * time for an object that is traveling at constant initial velocity. But since the object is accelerating, it will go an EXTRA distance. So we have to add the "0.5at²" in.

(3) d = vf*t - 0.5at²

"vf*t" would be the displacement if the object was traveling at the final velocity the entire time. But it didn't, it traveled at a slower velocity and accelerated. So we have to subtract some displacement because it wasn't going that fast the entire time. Thus, we have to subtract "0.5at²" out.

(4) d = 0.5*(vo + vf)*t

What is "0.5(vo + vf)"? That's the average velocity the object is traveling. So this equation really says "displacement = average velocity * time" which is an equation you already know.

(5) d = 0.5*(vf² - vo²)/a

This is just the restatement of equation (4). Multiplying the average velocity by time gives us the displacement. In this case, time = (vf - vo)/a which simplifies into this.

By understanding where the equations come from and what the meaning is behind each equation, you not only have an easier time remembering each equation, but you have the deeper understanding required to do well on the MCAT.

(Created with MS Paint)

Last edited:

- Nov 6, 2007

- 8,810

- 163

- 37

- Resident [Any Field]

Both and .Not to hijack the thread and change course here, but I scanned your med apps Geekchick and all I can think is "are the adcom people at Jefferson insane or stupid?" How did they waitlist you with your well-rounded, star-studded profile? Amazing in every category! Maybe they did that to see how serious you are about their school, figuring you'd follow up and contact them if you were serious. I just don't get what they could be thinking.

- We now return you to your regularly scheduled thread

Thank you, Berk! I wish I knew. I think I FUBARed the interview. It was my first and I really like Jeff and I think I just let my nerves and inexperience with the process get the better of me. There is supposedly a fair amount of movement on the HP waitlist, though. I told them I still wanted to be considered as Jeff's always been my first choice (which is true, though I also really liked Temple a lot), and I got some encouraging words from the Dean of Admissions about my chances in response. I will start stalking them again in early 2011 when it typically starts moving.

tl;dr i'm going to stick with calculusUgh okay guys, kinematics lesson.

Deriving the kinematics equations from the definition of uniform accelerationdoesn't require calculus at all.I recommend every MCAT student to derive them themselves so they can memorize them forever.Of course do this before the MCAT.It takes maybe 20 minutes.

vf = final velocity, vo = initial velocity

Let's start with the definition of acceleration: vf = vo + at (this one you have to learn by rote memorization). Since this is a linear equation, we can graph this. You should know that the slope of this graph is the acceleration and the area under the curve is displacement (d).

(1) vf = vo + at

Now let's figure out how we can use geometry to figure out the area to get displacement.

The area of the bottom rectangle is height*base or vo*t.

The area of each triangle is 0.5*height*base or 0.5*at*t or 0.5at².

The area of the overall rectangle (from the top red line to the time axis) is height*base or vf*t.

The area under the curve is therefore:

the bottom rectangle + the bottom triangle or

(2) d = vo*t + 0.5at²

the overall rectangle - the top triangle or

(3) d = vf*t - 0.5at²

Now what if we chop the top half of the triangle and put on the other side? We would get a rectangle whose area would be equal to the original area (see picture below).

The area of this rectangle is height*base or

(4) d = 0.5*(vo + vf)*t

The last equation is tricky. We have to plug in the first equation into the fourth equation. From equation (1), we rearrange tot = (vf - vo)/aand plug t into equation (4).

d = 0.5*(vo + vf)*(vf - vo)/aor

(5) d = 0.5*(vf² - vo²)/a

Let's look at the MEANING behind each equation.

(1) vf = vo + at

This is the definition of acceleration. It means, that acceleration is equal to the change in velocity over time.

(2) d = vo*t + 0.5at²

What if we take out "0.5at²"? What does "vo*t" physically mean? That is just distance = rate * time for an object that is traveling at constant initial velocity. But since the object is accelerating, it will go an EXTRA distance. So we have to add the "0.5at²" in.

(3) d = vf*t - 0.5at²

"vf*t" would be the displacement if the object was traveling at the final velocity the entire time. But it didn't, it traveled at a slower velocity and accelerated. So we have to subtract some displacement because it wasn't going that fast the entire time. Thus, we have to subtract "0.5at²" out.

(4) d = 0.5*(vo + vf)*t

What is "0.5(vo + vf)"? That's the average velocity the object is traveling. So this equation really says "displacement = average velocity * time" which is an equation you already know.

(5) d = 0.5*(vf² - vo²)/a

This is just the restatement of equation (4). Multiplying the average velocity by time gives us the displacement. In this case, time = (vf - vo)/a which simplifies into this.

By understanding where the equations come from and what the meaning is behind each equation, you not only have an easier time remembering each equation, but you have the deeper understanding required to do well on the MCAT.

(Created with MS Paint)

Took longer than 20 mins just to read this post. Just memorize the damned kinematics equations. If you cannot make the concepts that they describe make sense to the point where deriving them is your only option, instead of memorizing 3-5 simple equations (depending on prep company), feel free to find another line of work. The "rocket science" on the exam (projectile motion) isn't exactly rocket science.

Ugh okay guys, kinematics lesson.

Deriving the kinematics equations from the definition of uniform accelerationdoesn't require calculus at all.I recommend every MCAT student to derive them themselves so they can memorize them forever.Of course do this before the MCAT.It takes maybe 20 minutes.

....

(Created with MS Paint)

- May 25, 2007

- 3,992

- 806

The "rocket science" on the exam (projectile motion) isn't exactly rocket science.

Now that's the funniest thing I've read in a long time.

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